My Physics teacher stated that Pluto has a gravitational pull on objects on Earth, namely humans. Is this true? What is the free-fall acceleration of Pluto with respect to being on the Earth's surface? (i.e. the Earth's free-fall acceleration is 9.8 m/s*s).

3 Answers
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This is true. Newton's law of universal gravitation says everything attracts everything. To get the free-fall acceleration of some object on Earth towards Pluto, take Newton's law and divide by the object's mass to get $$a=\frac{F}{m}=\frac{GM}{r^2}.$$ Subbing in reasonable values - $M=0.002$ Earth masses, and $r$ between 29 and 49 AU you get something like $10^{-14}\textrm{ m s}^{-2}$.

I am not a physicist but this will answer your query; According to Newton's law of universal gravitation every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$F=G\frac{m_{1}m_{2}}{r^{2}}$

for example:
lets say r is the distance between earth and pluto, G the universal gravitation constant, m1 is the mass of pluto and m2 be the mass of a human on earth, we can then calculate the gravitational force exerted by m1 to m2.

As Emilio points out the gravitational attraction by pluto is on the order of $10^{-14}\textrm{ m s}^{-2}$.

However this is not an acceleration that can be felt in any meaningful way - In a flat gravitational field movement due to acceleration by gravity is identical to inertial movement. However as it happens the gravitational field from pluto is not flat, it's inversely proportional to the square of the distance. This means that with a suitable reference point it's in theory possible to detect the tidal force that pluto creates.

For us a suitable reference point would be earth.

We know, from the other answers, that the acceleration from pluto's pull is
$$a_{me}=\frac{GM}{d^2}.$$

Where $M$ is the mass of Pluto and $d$ is our distance to Pluto.

If we assume that we are on the side of the planet facing Pluto, and on a direct line between the center of the planet and the plutoid. We get that the acceleration of earth due to Pluto is
$$a_{earth}=\frac{GM}{(d+r)^2}.$$

Where $r$ is the radius of the earth.

The difference is thus
$$a_{me}-a_{earth}=\frac{GM}{d^2}-\frac{GM}{(d+r)^2}=GMr(\frac{2}{d^3+d^2r}-\frac{r}{d^4+2d^3r+d^2r^2})$$

However the radius of earth is tiny compared to the distance to Pluto so we can approximate the above to;
$$a_{me}-a_{earth}\approx\frac{2GMr}{d^3}$$

Pluging this into WolframAlpha tells us that we can perceive a difference in acceleration of roughly 10-19ms-2 between us and the Earth.
For reference this is about 7 Angstrom per day per day. Or 2 inches pr second in the current age of the universe.
How to actually messure accelerations of those magnitudes are left as an exercise for the reader.

This is, of course, the tidal force exerted by Pluto on the Earth. I think the best expression for this acceleration is one Angstrom per day per day.
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Emilio PisantyAug 28 '14 at 13:13

@EmilioPisanty Answer updated. Though I must say that I quite like 6qdr Planck lengths per second per second. 7 Angstrom per day per day just doesn't have the same ring to it.
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TaemyrAug 28 '14 at 13:37

I find both 'quadrillion' and 'Planck length' to be truly inscrutable terms, to be honest. If you're just going to pile zeroes, then why not say one quintillionth of a meter per second${}^2$? On the other hand, I have some vague (and hard-earned) understanding of just how small an Angstrom is, so I can just about comprehend, I think, 1A/day${}^2$.
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Emilio PisantyAug 28 '14 at 13:44

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@EmilioPisanty I did not say that quadrillions planck lengths per second per second is a good unit to measure anything in - note that I did update the answer to your suggestion. As an accesible value I think 2 inches per second per current age of the universe is about the best one can have.
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TaemyrAug 28 '14 at 13:49